Question

If α,β are zeroes of the quadratic polynomial x2 + 2x - 1, then find the value of 1/αβ² + 1/α²β​

Answer 1

Answer:

$\bf{ \dfrac{1}{ { \alpha }^{2} \beta}} + \dfrac{1}{ { \alpha { \beta}}^{2}} = - 2$

Step-by-step explanation:

Polynomial = x² + 2x – 1

Value of $\sf{ \dfrac{1}{ { \alpha }^{2} \beta}} + \dfrac{1}{ { \alpha { \beta}}^{2}}$ = ??

We know that,

$\sf{\alpha + \beta = \dfrac{ - b}{a}}$ and $\sf{\alpha \beta = \dfrac{c}{a}}$

In the polynomial,

• a = 1
• b = 2
• c = -1

$\longrightarrow \: \sf{\alpha + \beta = \dfrac{ - b}{a}}$

$\longrightarrow \: \sf{\alpha + \beta = \dfrac{ -2}{1}}$

$\longrightarrow \: \sf{\alpha + \beta = - 2}$

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$\longrightarrow \: \sf{\alpha \beta = \dfrac{ c}{a}}$

$\longrightarrow \: \sf{\alpha \beta = \dfrac{ - 1}{1}}$

$\longrightarrow \: \sf{\alpha \beta = - 1}$

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★ Value of $\sf{ \dfrac{1}{ { \alpha }^{2} \beta}} + \dfrac{1}{ { \alpha { \beta}}^{2}}$ :

$\longrightarrow \: \sf{ \dfrac{1}{ { \alpha }^{2} \beta}} + \dfrac{1}{ { \alpha { \beta}}^{2}}$

$\longrightarrow \: \sf{ \bigg(\dfrac{1}{ { \alpha }^{2} \beta} \times \dfrac{ \beta }{ \beta }\bigg)} + \bigg(\dfrac{1}{ { \alpha { \beta}}^{2}} \times \dfrac{ \alpha }{ \alpha } \bigg)$

$\longrightarrow \: \sf{ \dfrac{ \beta }{ { \alpha }^{2} \beta^{2} }} + \dfrac{ \alpha }{ { \alpha^{2} { \beta}}^{2}}$

$\longrightarrow \: \sf{ \dfrac{ \alpha + \beta }{ { \alpha }^{2} \beta^{2} }}$

$\longrightarrow \: \sf{ \dfrac{ - 2 }{ (\alpha \beta)^{2} }}$

$\longrightarrow \: \sf{ \dfrac{ - 2 }{ ( - 1)^{2} }}$

$\longrightarrow \: \sf{ \dfrac{ - 2 }{1 }}$

$\longrightarrow \:\sf{ - 2}$

Value of $\sf{ \dfrac{1}{ { \alpha }^{2} \beta}} + \dfrac{1}{ { \alpha { \beta}}^{2}}$ is –2.

$\bf{\therefore \: \dfrac{1}{ { \alpha }^{2} \beta}} + \dfrac{1}{ { \alpha { \beta}}^{2}} = - 2$